Irrational numbers - incomprehnsible?

[WW_III_ANGRY]

If "it" means the approximations, then no, they are never exact. But \sqrt{2} is the exact value of the number whose square is 2.
That's not true. Some of the approximations give values that are larger than 2. For example, 1.414214² = 2.000001238 by my calculator.
.

Actually the square root is 1.414213 yes? Which I still show as 1.9999... (with some string of other integers dependent upon how far out we go at the tail end)

 

[Mark44]

Actually the square root is 1.414213 yes?

No. I rounded up, and you rounded down. sqrt(2) is between 1.414213 and 1.414214.

Which I still show as 1.9999... (with some string of other integers dependent upon how far out we go at the tail end)

 

[WW_III_ANGRY]

No. I rounded up, and you rounded down. sqrt(2) is between 1.414213 and 1.414214.

Oh I see. I didn't wish to round, I terminated the sequence without rounding.

 

[JSuarez]

however that is an incomprehensible value to me as well as it is unable to be expressed in the physical, or conceptually through decimal expansion or fraction. Which is why I see it as irrational in the whole sense of the word :)

But why should these be the only possible ways of "comprehending" that particular root? Consider the sequence:

x_0=1
x_{n+1}=\frac{2-{x^2}_n}{2x_n}

It has a limit, equal to the positive root of x^2-2 and both these facts may be proven independently; moreover, if you multiply it termwise by itself, you'll get again a convergent sequence, whose limit is 4.

 

[jbunniii]

As I understand it also, finding the numerical value in decimals is the most precise method to evaluate the square root of 2. If there is another more precise method can you please let me know? Thanks,

I think you are conflating two different issues.

Not every real number has a decimal expansion with finitely many digits. In fact, most do not.

But this is not unique to irrational numbers. There are many rational numbers that do not have terminating decimal expansions. 1/3 is an example.

That \sqrt{2} is irrational means precisely that it cannot be expressed as a ratio of two integers: there are no m,n \in \mathbb{Z} such that

\left(\frac{m}{n}\right)^2 = 2.

This shows that 2 does not have a square root in the rational field \mathbb{Q}. The existence of such gaps in \mathbb{Q} is the motivating factor behind the construction of the larger field \mathbb{R} (real numbers), which does not have such gaps.

There are several well-defined ways to construct \mathbb{R} from \mathbb{Q}. One standard method uses Cauchy sequences, and another one uses Dedekind cuts. Just about any real analysis book should contain at least one of these constructions. Some calculus books do, too. (E.g., Spivak.)

 

[Mark44]

Mark, thank you for bearing with me :)
Let me continue with my understanding in hope that you can see what I am thinking and where I am wrong:

1/3 is able to be represented precisely in a fraction form. Essentially its a piece divided into three parts. Makes perfect logical sense to me and is comprehensible as well as rational and a rational number. I don't see how the comparison as applicable because my dilemma is with a square root which is reducible to fractions or decimals in rational numbers and is also comprehensible and rational to do so if needed. As I see it everything we know of physically can be divided as fractions or decimal expansions, or what have you. As I conceptualize square roots these too must follow the same path logically or in the case of irrational numbers these aren't applicable.

OK, we are able to divide a whole (1) into three equal parts that add to 1. IOW 1 = 1/3 + 1/3 + 1/3. Sounds like you're comfortable with that, even though when you convert the fractions to decimal representations, you get something shy of 1.

Let's take it a step further. With \sqrt{2}, we are dividing 2 into two equal parts that multiply to make 2. IOW, \sqrt{2} * \sqrt{2} = 2. The main difference here, as I see it, is that the decimal representations are not easily predictable. For example, it's clear that the digit in the 25th decimal place of the decimal representation of 1/3 is 3, but although I can find the digit in the 25th decimal place of \sqrt{2}, it will take me a bit of work and a lot of paper to do it.

However precise a decimal representation of \sqrt{2} you need, I can find enought decimal digits of \sqrt{2} to satisfy that precision. If we're talking about light years, as you were in a previous thread, all I need is about 13 or 14 digits of \sqrt{2} to have a precision within a mile. If you need more precision than that, I'll just do some more calculations and get more digits.

What's more, using nothing more than paper and pencil, I can supply you an approximation of \sqrt{2} that is so close to the actual value that if you squared it on your calculator (assuming you have one), the result would be 2. That's just plain 2, with no decimal point and no digits after the 2. I would have "fooled" your calculator by giving it a value that's not quite equal to \sqrt{2}. The calculator would multiply the value I supply by itself and come up with a result so close to 2, that the calculator would be unable to distinguish the result from 2.

I understand geometrically that 1/3 is more desirable than .333, however as stated above this is able to be comprehended conceptually and physically. I understand the square root of 2 is able to be utilized to find the hypotenuse of a right triangle, however that is an incomprehensible value to me as well as it is unable to be expressed in the physical, or conceptually through decimal expansion or fraction. Which is why I see it as irrational in the whole sense of the word :)

1/3 is more desirable computationally as well, since using .333 in place of 1/3 results in errors. If you want smaller errors, use a representation that is more precise. Same with using a decimal representation of \sqrt{2}: the more digits in your representation, the better your calculated result. What's so incomprehensible about that?

Even ordinary decimal fractions don't always work the way we think they should when we start doing calculations with calculators and computers. For example, 1/10 = 0.1 exactly, but in some programming languages (such as C and languages based on C), if you add this number to itself 100 times (meaning that you have an addition problem with 100 terms, all of which are .1), you don't get exactly 10. The reason for this is that 1/10 has a very nice, short decimal representation in base 10, but computers usually work in different number systems, base 2 or base 16, and conversion from one base to another can lead to surprising results. So how comprehensible is that?

The upshot of my rambling here is that a significant chunk of mathematics is concerned with approximations, and how close the approximations are to what they are supposed to represent. Like or or not, most numbers (by far) are irrational, so the best we can do when we need to calculate something with them is to use as much precision in our representations as is required to get the results we're after.

 

[WW_III_ANGRY]

OK, we are able to divide a whole (1) into three equal parts that add to 1. IOW 1 = 1/3 + 1/3 + 1/3. Sounds like you're comfortable with that, even though when you convert the fractions to decimal representations, you get something shy of 1.

Let's take it a step further. With \sqrt{2}, we are dividing 2 into two equal parts that multiply to make 2. IOW, \sqrt{2} * \sqrt{2} = 2. The main difference here, as I see it, is that the decimal representations are not easily predictable. For example, it's clear that the digit in the 25th decimal place of the decimal representation of 1/3 is 3, but although I can find the digit in the 25th decimal place of \sqrt{2}, it will take me a bit of work and a lot of paper to do it.

However precise a decimal representation of \sqrt{2} you need, I can find enought decimal digits of \sqrt{2} to satisfy that precision. If we're talking about light years, as you were in a previous thread, all I need is about 13 or 14 digits of \sqrt{2} to have a precision within a mile. If you need more precision than that, I'll just do some more calculations and get more digits.

What's more, using nothing more than paper and pencil, I can supply you an approximation of \sqrt{2} that is so close to the actual value that if you squared it on your calculator (assuming you have one), the result would be 2. That's just plain 2, with no decimal point and no digits after the 2. I would have "fooled" your calculator by giving it a value that's not quite equal to \sqrt{2}. The calculator would multiply the value I supply by itself and come up with a result so close to 2, that the calculator would be unable to distinguish the result from 2.


1/3 is more desirable computationally as well, since using .333 in place of 1/3 results in errors. If you want smaller errors, use a representation that is more precise. Same with using a decimal representation of \sqrt{2}: the more digits in your representation, the better your calculated result. What's so incomprehensible about that?

Even ordinary decimal fractions don't always work the way we think they should when we start doing calculations with calculators and computers. For example, 1/10 = 0.1 exactly, but in some programming languages (such as C and languages based on C), if you add this number to itself 100 times (meaning that you have an addition problem with 100 terms, all of which are .1), you don't get exactly 10. The reason for this is that 1/10 has a very nice, short decimal representation in base 10, but computers usually work in different number systems, base 2 or base 16, and conversion from one base to another can lead to surprising results. So how comprehensible is that?

The upshot of my rambling here is that a significant chunk of mathematics is concerned with approximations, and how close the approximations are to what they are supposed to represent. Like or or not, most numbers (by far) are irrational, so the best we can do when we need to calculate something with them is to use as much precision in our representations as is required to get the results we're after.

That is great, very comprehensible thank you.

But what does this mean about the decimal system, is it flawed or is the square root of 2 flawed? Is it possible to postulate a more precise number system? I consider math to be "perfect" conceptually, even though that requirement for squaring two is not necessary as you elegantly put above.

 

[WW_III_ANGRY]

I think you are conflating two different issues.

Not every real number has a decimal expansion with finitely many digits. In fact, most do not.

But this is not unique to irrational numbers. There are many rational numbers that do not have terminating decimal expansions. 1/3 is an example.

That \sqrt{2} is irrational means precisely that it cannot be expressed as a ratio of two integers: there are no m,n \in \mathbb{Z} such that

\left(\frac{m}{n}\right)^2 = 2.

This shows that 2 does not have a square root in the rational field \mathbb{Q}. The existence of such gaps in \mathbb{Q} is the motivating factor behind the construction of the larger field \mathbb{R} (real numbers), which does not have such gaps.

There are several well-defined ways to construct \mathbb{R} from \mathbb{Q}. One standard method uses Cauchy sequences, and another one uses Dedekind cuts. Just about any real analysis book should contain at least one of these constructions. Some calculus books do, too. (E.g., Spivak.)

Thanks I'm aware of Dedekind cuts but not Cauchy sequences. I am really low level math.

 

[Office_Shredder]

But what does this mean about the decimal system, is it flawed or is the square root of 2 flawed? Is it possible to postulate a more precise number system? I consider math to be "perfect" conceptually, even though that requirement for squaring two is not necessary as you elegantly put above.

There's a way of comparing sizes of infinity in which you can prove that you can't list every single real number between zero and one. If there was a number system in which every real number between zero and one had a finite decimal expansion, you would be able to list all of them (first write down every one with a single decimal digit, then all of them with two decimals, then all of them with three decimals, etc. Every number will be written down eventually).

So the upshot is that it's not a flaw of anything, it's just the way that math works

 

[jbunniii]

That is great, very comprehensible thank you.

But what does this mean about the decimal system, is it flawed or is the square root of 2 flawed? Is it possible to postulate a more precise number system? I consider math to be "perfect" conceptually, even though that requirement for squaring two is not necessary as you elegantly put above.

I think the answer to your question is no, it's not possible.

If I may take the liberty of interpreting your question, the basic issue is this: is there any scheme by which EVERY real number (or geometrically, the coordinate of every point on a line) can be represented using a finite number of integers? [Any rational number, for example, can be represented using two integers. One might hope that by using more integers, we might be able to cover the whole real line.]

The answer to that question is no, and the reason is that the cardinalities are not the same.

If \mathbb{Z} is the set of all integers, and more generally \mathbb{Z}^n is the set of all vectors of n integers, then I can define the set of all "representations using a finite number of integers" to be

S = \bigcup_{n=1}^{\infty} \mathbb{Z}^n

But this is a countable union of countable sets, and hence S itself is countable. On the other hand there are uncountably many real numbers (or equivalently, uncountably many points on a line), so there's no hope to represent all of them using such a system. In fact, any such system will represent "almost none" of them! (The set will of representable numbers/points will have measure 0.)

[Edit]: I just saw Office_Shredder's post, which is saying the same thing as mine but with less technical jargon.